Answer:
D. There is a reduced use of pesticides, herbicides, and fertilizers
Explanation:
Answer:
(a) 1/2; (b) no
Explanation:
Glucose-6-phosphate dehydrogenase deficiency (G6PD) is an X-linked recessive disorder and the woman's father was diseased so it means that woman is a carrier of the allele but has normal phenotype. It means that she will have XXᵇ genotype.
In contrast to this, her husband is diseased so his genotype will be XᵇY.
The Punnett square diagram related to the cross is attached.
(a) Proportion of their sons expected to be G6PD is 1/2:
They both may give birth to 4 progeny with genotypes XXᵇ, XᵇXᵇ, XY and XᵇY. It means they both may have 2 sons out of which one with genotype XᵇY will be diseased while the one with genotype XY will be healthy. So the proportion of their sons having G6PD is 1/2 or 50%.
(b) If the husband were G6PD deficient, the answer will not change.
The reason behind this is that this disease is caused by an allele located in X chromosome. But father contributes only Y chromosome to his son not X chromosome. The X chromosome will affect the genotype of his daughter not son that is why answer will not change. It means they will still have 1/2 of their sons diseased.
<span>He is at an increased risk for sudden infant death. Since he is sleeping on a mattress that might inhibit his breathing, as well as having respiratory issues that might do the same (along with parents who are smoking, which will only inhibit proper respiration further), he will need to be constantly checked to make sure he is sleeping properly.</span>
Geo wild card spot on my a open relationship between the two of you are not up for
Answer:
This is due to the event of Speciation that happened for the rodents in Island B but not for the rodents in Island C.
Explanation:
- Due to splitting of the population,
- The sub-population of rodents formed in Island B are B1 and B2.
- The sub-population of rodents formed in Island C are C1 and C2.
- In case of Island B, each of the B1 and B2 sub-populations that got split from each other developed certain mutations that were necessary for them to adapt to the particular diverse environment each of them were exposed to, through the period of 50,000 years. These mutations were so varied that reproductive isolation was generated between them that resulted in each of them to develop into different species.Hence, speciation happens here and B1 and B2 are incapable of inter-breeding.
- In case of Island C, each of the C1 and C2 sub-populations that got split might have got exposed to similar environmental change or no environmental change or the environmental change might have been too small to cause drastic change in each of the sub-populations. As a result of this the two sub-populations might have acquired certain mutations to adapt to the environment each of them were exposed to, through a period of 100,000 years. These mutations might not have been too variable or contrasting to cause reproductive isolation between C1 and C2. Hence, no new speciation happens here and C1 and C2 are capable of inter-breeding.
